Evaluate Powers of Complex Numbers Using DeMoivre's Theorem - Pre-Calculus

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Find the magnitude of the complex number described by .

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To find the magnitude of a complex number we use the formula:

,

where our complex number is in the form .

Therefore,

Find the magnitude of :

, where the complex number satisfies .

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Note for any complex number z, we have:

.

Let . Hence ${}|z|=$\sqrt{10}$$

Therefore:

This gives the result.

What is the magnitude of ?

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To find the magnitude of a complex number we use the following formula:

, where .

Therefore we get,

.

Now to find

.

Simplify

$\small (\cos x+i\sin $x)^3$$

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We can use DeMoivre's formula which states:

Now plugging in our values of $\small n=3$ and we get the desired result.

$\small (\cos x+i\sin $x)^n$=(\cos nx+i\sin nx)$

$\cos 3x +i \sin 3x$

Evaluate:

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First, convert this complex number to polar form.

$r= $\sqrt{ $a^2$ + $b^2$ }$ = $$\sqrt{1^2$ $+(-1)^2$ }$=$\sqrt{2}$$

$\sin \theta = $\frac{-1}{\sqrt2}$$

$\theta = $\frac{5 \pi }{4}$ \enspace or \enspace $\frac{7 \pi }{4}$$

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is $\frac{7\pi}{4}$ .

This gives us $($\sqrt{2}$ ( \cos $\frac{7 \pi }{4}$ + i \sin $\frac{ 7 \pi }{4}$ )) ^ 8$

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$(\sqrt2 $)^8$ ( \cos $\frac{ 7 \pi }{4}$ \cdot 8 + i \sin $\frac{ 7 \pi }{4}$ \cdot 8 )$ simplifying

$2 ^ 4 ( \cos 14 \pi + i \sin 14 \pi )$ , $14 \pi$ is coterminal with since it is an even multiple of $\pi$

$2^ 4 (\cos 0 + i \sin 0 ) = 16(1 + 0i) = 16$

Use DeMoivre's Theorem to evaluate the expression $(\sqrt3 - 3i ) ^ 6$ .

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First convert this complex number to polar form:

$r = $\sqrt{ $(\sqrt3)^2$ + $1^2$ }$ = $\sqrt{ 3 + 1 }$ = \sqrt4 = 2$

$\sin \theta = $\frac{ 1}{ 2}$$ so $\theta = $\frac{\pi }{6}$ \enspace or \enspace $\frac{5 \pi }{6}$$

Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is $\frac{ \pi }{6}$

So we are evaluating $(2(\cos $\frac{ \pi }{6}$ + i \sin $\frac{ \pi }{6}$ $))^3$$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$(3 \sqrt3 -3i) ^ 6$

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First convert this point to polar form:

$r = $$\sqrt{(3\sqrt3)^2$ + $(-3)^2$ }$ = $\sqrt{9 \cdot 3 + 9 }$ = $\sqrt{36}$ = 6$

$\cos \theta = $\frac{ 3\sqrt3}{6 }$ = $\frac{\sqrt3}{2}$$

$\theta = \cos ^{-1} ($\frac{\sqrt3}{2}$ ) = $\frac{ \pi }{6}$ \enspace \or \enspace $\frac{ 11 \pi }{6}$$

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is $\frac{ 11 \pi }{6}$

We are evaluating $(6 \cos $\frac{11 \pi }{6}$ + i \sin $\frac{ 11 \pi }{6}$ ) ^ 6$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$$\frac{11 \pi }{6}$ \cdot 6 = 11 \pi$ which is coterminal with $\pi$ since it is an odd multiplie

$46,656(\cos \pi + i \sin \pi ) = 46,656 (-1 + 0i) = -46,656$

Evaluate $(2 -2 $\sqrt3)^6$$

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First, convert this complex number to polar form:

$r = $$\sqrt{2^2$ + $(2\sqrt3)^2$ }$ = $\sqrt{4 +12}$ = $\sqrt{16}$ = 4$

$\sin \theta = $\frac{ -2\sqrt3 }{4}$ = -$\frac{\sqrt3}{2}$$

$\theta = \sin ^{-1} (-$\frac{\sqrt3}{2}$) = $\frac{4 \pi }{3}$ \enspace or \enspace $\frac{ 5 \pi }{3}$$

Since the real part is positive and the imaginary part is negative, this is in quadrant IV, so the angle is $\frac{5 \pi }{3}$

So we are evaluating $(4(\cos $\frac{5 \pi }{3}$ + i \sin $\frac{5 \pi }{3}$ ) $)^6$$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$10 \pi$ is coterminal with since it is an even multiple of $\pi$

$4,096( \cos 0 + i \sin 0 ) = 4,096 ( 1 + 0i) = 4,096$

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First, convert the complex number to polar form:

$r = \sqrt{ 2^2 + 2^2 } = \sqrt{4 + 4 } = \sqrt{2 \cdot 4 } = 2 \sqrt 2$

$\sin \theta = \frac{ 2 }{ 2 \sqrt 2 } = \frac{ 1 }{ \sqrt2 }$

$\theta = \sin ^{-1 } (\frac{ 1}{ \sqrt 2 }) = \frac{ \pi }{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is $\frac{ \pi }{4}$

This means we're evaluating

$(2 \sqrt2 (\cos \frac{ \pi }{4} + i \sin \frac{ \pi }{4} ))^ 9$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(2 \sqrt2 )^9 (\cos \frac{\pi }{4} \cdot 9 + i \sin \frac{ \pi }{4} \cdot 9 )$

First, evaluate $(2\sqrt2)^9$ . We can split this into $2^9 \cdot (\sqrt2)^9$ which is equivalent to $2^9 \cdot (\sqrt2)^8 \cdot \sqrt2 = 2^9 \cdot 2^4 \cdot \sqrt 2$

\[We can re-write the middle exponent since $\sqrt 2$ is equivalent to $2^{\frac{1}{2}}$ \]

This comes to $8,192 \sqrt 2$

Evaluating sine and cosine at $\frac{\pi }{4} \cdot 9 = \frac{ 9 \pi }{4}$ is equivalent to evaluating them at $\frac{ \pi }{4}$ since $\frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$

This means our expression can be written as:

$8,192 \sqrt2 ( \cos \frac{\pi }{4} + i \sin \frac{ \pi }{4} ) = 8,192 \sqrt2 (\frac{1}{\sqrt2} + i \frac{ 1}{\sqrt2}) = 8,192 + 8,192i$

Evaluate $(-3 + 3i ) ^{ 12}$

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First convert the complex number into polar form:

$r = $$\sqrt{3^2$ + $(-3)^2$ }$ = $\sqrt{9+9}$ = $\sqrt{2 \cdot 9 }$ = 3 \sqrt2$

$\sin \theta = $\frac{3}{3 \sqrt2}$ = $\frac{ 1}{\sqrt2}$$

$\theta = \sin ^{-1 } ($\frac{1}{\sqrt2}$) = $\frac{\pi}{4}$ \enspace or \enspace $\frac{ 3 \pi }{4}$$

Since the real part is negative but the imaginary part is positive, the angle should be in quadrant II, so it is $\frac{ 3 \pi }{4}$

We are evaluating $(3\sqrt2 (\cos $\frac{ 3 \pi }{4}$ + i \sin $\frac{ 3 \pi }{4}$ ) ) ^ {12}$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

simplify and take the exponent

$34,012,224 (\cos 9\pi + i \sin 9 \pi )$

$9\pi$ is coterminal with $\pi$ since it is an odd multiple of pi

$34, 012, 224 ( \cos \pi + i \sin \pi ) = 34, 012, 224 ( -1 + 0i) = -34,012, 224$

Find the magnitude of the complex number described by .

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To find the magnitude of a complex number we use the formula:

,

where our complex number is in the form .

Therefore,

Find the magnitude of :

, where the complex number satisfies .

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Note for any complex number z, we have:

.

Let . Hence ${}|z|=$\sqrt{10}$$

Therefore:

This gives the result.

What is the magnitude of ?

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To find the magnitude of a complex number we use the following formula:

, where .

Therefore we get,

.

Now to find

.

Simplify

$\small (\cos x+i\sin $x)^3$$

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We can use DeMoivre's formula which states:

Now plugging in our values of $\small n=3$ and we get the desired result.

$\small (\cos x+i\sin $x)^n$=(\cos nx+i\sin nx)$

$\cos 3x +i \sin 3x$

Evaluate:

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First, convert this complex number to polar form.

$r= $\sqrt{ $a^2$ + $b^2$ }$ = $$\sqrt{1^2$ $+(-1)^2$ }$=$\sqrt{2}$$

$\sin \theta = $\frac{-1}{\sqrt2}$$

$\theta = $\frac{5 \pi }{4}$ \enspace or \enspace $\frac{7 \pi }{4}$$

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is $\frac{7\pi}{4}$ .

This gives us $($\sqrt{2}$ ( \cos $\frac{7 \pi }{4}$ + i \sin $\frac{ 7 \pi }{4}$ )) ^ 8$

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$(\sqrt2 $)^8$ ( \cos $\frac{ 7 \pi }{4}$ \cdot 8 + i \sin $\frac{ 7 \pi }{4}$ \cdot 8 )$ simplifying

$2 ^ 4 ( \cos 14 \pi + i \sin 14 \pi )$ , $14 \pi$ is coterminal with since it is an even multiple of $\pi$

$2^ 4 (\cos 0 + i \sin 0 ) = 16(1 + 0i) = 16$

Use DeMoivre's Theorem to evaluate the expression $(\sqrt3 - 3i ) ^ 6$ .

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First convert this complex number to polar form:

$r = $\sqrt{ $(\sqrt3)^2$ + $1^2$ }$ = $\sqrt{ 3 + 1 }$ = \sqrt4 = 2$

$\sin \theta = $\frac{ 1}{ 2}$$ so $\theta = $\frac{\pi }{6}$ \enspace or \enspace $\frac{5 \pi }{6}$$

Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is $\frac{ \pi }{6}$

So we are evaluating $(2(\cos $\frac{ \pi }{6}$ + i \sin $\frac{ \pi }{6}$ $))^3$$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$(3 \sqrt3 -3i) ^ 6$

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First convert this point to polar form:

$r = $$\sqrt{(3\sqrt3)^2$ + $(-3)^2$ }$ = $\sqrt{9 \cdot 3 + 9 }$ = $\sqrt{36}$ = 6$

$\cos \theta = $\frac{ 3\sqrt3}{6 }$ = $\frac{\sqrt3}{2}$$

$\theta = \cos ^{-1} ($\frac{\sqrt3}{2}$ ) = $\frac{ \pi }{6}$ \enspace \or \enspace $\frac{ 11 \pi }{6}$$

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is $\frac{ 11 \pi }{6}$

We are evaluating $(6 \cos $\frac{11 \pi }{6}$ + i \sin $\frac{ 11 \pi }{6}$ ) ^ 6$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$$\frac{11 \pi }{6}$ \cdot 6 = 11 \pi$ which is coterminal with $\pi$ since it is an odd multiplie

$46,656(\cos \pi + i \sin \pi ) = 46,656 (-1 + 0i) = -46,656$

Evaluate $(2 -2 $\sqrt3)^6$$

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First, convert this complex number to polar form:

$r = $$\sqrt{2^2$ + $(2\sqrt3)^2$ }$ = $\sqrt{4 +12}$ = $\sqrt{16}$ = 4$

$\sin \theta = $\frac{ -2\sqrt3 }{4}$ = -$\frac{\sqrt3}{2}$$

$\theta = \sin ^{-1} (-$\frac{\sqrt3}{2}$) = $\frac{4 \pi }{3}$ \enspace or \enspace $\frac{ 5 \pi }{3}$$

Since the real part is positive and the imaginary part is negative, this is in quadrant IV, so the angle is $\frac{5 \pi }{3}$

So we are evaluating $(4(\cos $\frac{5 \pi }{3}$ + i \sin $\frac{5 \pi }{3}$ ) $)^6$$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

$10 \pi$ is coterminal with since it is an even multiple of $\pi$

$4,096( \cos 0 + i \sin 0 ) = 4,096 ( 1 + 0i) = 4,096$

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First, convert the complex number to polar form:

$r = \sqrt{ 2^2 + 2^2 } = \sqrt{4 + 4 } = \sqrt{2 \cdot 4 } = 2 \sqrt 2$

$\sin \theta = \frac{ 2 }{ 2 \sqrt 2 } = \frac{ 1 }{ \sqrt2 }$

$\theta = \sin ^{-1 } (\frac{ 1}{ \sqrt 2 }) = \frac{ \pi }{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is $\frac{ \pi }{4}$

This means we're evaluating

$(2 \sqrt2 (\cos \frac{ \pi }{4} + i \sin \frac{ \pi }{4} ))^ 9$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(2 \sqrt2 )^9 (\cos \frac{\pi }{4} \cdot 9 + i \sin \frac{ \pi }{4} \cdot 9 )$

First, evaluate $(2\sqrt2)^9$ . We can split this into $2^9 \cdot (\sqrt2)^9$ which is equivalent to $2^9 \cdot (\sqrt2)^8 \cdot \sqrt2 = 2^9 \cdot 2^4 \cdot \sqrt 2$

\[We can re-write the middle exponent since $\sqrt 2$ is equivalent to $2^{\frac{1}{2}}$ \]

This comes to $8,192 \sqrt 2$

Evaluating sine and cosine at $\frac{\pi }{4} \cdot 9 = \frac{ 9 \pi }{4}$ is equivalent to evaluating them at $\frac{ \pi }{4}$ since $\frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$

This means our expression can be written as:

$8,192 \sqrt2 ( \cos \frac{\pi }{4} + i \sin \frac{ \pi }{4} ) = 8,192 \sqrt2 (\frac{1}{\sqrt2} + i \frac{ 1}{\sqrt2}) = 8,192 + 8,192i$

Evaluate $(-3 + 3i ) ^{ 12}$

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First convert the complex number into polar form:

$r = $$\sqrt{3^2$ + $(-3)^2$ }$ = $\sqrt{9+9}$ = $\sqrt{2 \cdot 9 }$ = 3 \sqrt2$

$\sin \theta = $\frac{3}{3 \sqrt2}$ = $\frac{ 1}{\sqrt2}$$

$\theta = \sin ^{-1 } ($\frac{1}{\sqrt2}$) = $\frac{\pi}{4}$ \enspace or \enspace $\frac{ 3 \pi }{4}$$

Since the real part is negative but the imaginary part is positive, the angle should be in quadrant II, so it is $\frac{ 3 \pi }{4}$

We are evaluating $(3\sqrt2 (\cos $\frac{ 3 \pi }{4}$ + i \sin $\frac{ 3 \pi }{4}$ ) ) ^ {12}$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

simplify and take the exponent

$34,012,224 (\cos 9\pi + i \sin 9 \pi )$

$9\pi$ is coterminal with $\pi$ since it is an odd multiple of pi

$34, 012, 224 ( \cos \pi + i \sin \pi ) = 34, 012, 224 ( -1 + 0i) = -34,012, 224$